1. ## Analytic Geometry Q10

Question:
Find the equation of the bisector of the pair of acute angles formed by the lines $\displaystyle L_1$ and $\displaystyle L_2$ given by $\displaystyle 4x-2y-10=0$ and $\displaystyle x-2y-1=0$, respectively.

Attempt:

$\displaystyle L_1 \rightarrow 4x - 2y - 15 = 0$
$\displaystyle A = 4, \ B = -2, \ C=-15$

$\displaystyle d_1 = \frac{Ax+By+C}{\pm\sqrt{A^2+B^2}} = \frac{4x-2y-15}{-\sqrt{4^2+2^2}} = \frac{4x-2y-15}{-2\sqrt{5}}$

$\displaystyle L_2 \rightarrow x - 2y - 1$
$\displaystyle A = 1, \ B = -2, \ C=-1$

$\displaystyle d_2 = \frac{Ax+By+C}{\pm\sqrt{A^2+B^2}} = \frac{x-2y-1}{-\sqrt{1^2+2^2}} = \frac{x-2y-1}{-\sqrt{5}}$

Now, I need to simplify the two distance equations to get the equation of the bisector...

$\displaystyle \frac{4x-2y-15}{-2\sqrt{5}} = \frac{x-2y-1}{-\sqrt{5}}$

Here is the problem! I don't know how to simplify with the square roots around , Need Help!

2. Originally Posted by looi76
[FONT="Georgia"][SIZE="3"]Question:
Find the equation of the bisector of the pair of acute angles formed by the lines $\displaystyle L_1$ and $\displaystyle L_2$ given by $\displaystyle 4x-2y-10=0$ and $\displaystyle x-2y-1=0$, respectively.

...
1. Both equations are given in normal form. The normal vectors are :

$\displaystyle \overrightarrow{n_1} = (4, -2)$ . Therefore the direction vector of the first line is $\displaystyle \vec u = (2, 4)$ with $\displaystyle |\overrightarrow{u}| = \sqrt{20}$

and

$\displaystyle \overrightarrow{n_2} = (1, -2)$ . Therefore the direction vector of the second line is $\displaystyle \vec v = (2,1)$ with $\displaystyle |\overrightarrow{v}| = \sqrt{5}$

2. The point of intersection is $\displaystyle P(3,1)$ . Therefore the angle bisector has to pass through this point too.

3. The direction of the angle bisector is the sum of the unit vectors of the two direction vectors:

$\displaystyle \vec b = \dfrac1{\sqrt{20}} (1,4) + \dfrac1{\sqrt{5}} (2,1) = \dfrac1{\sqrt{20}} (2,4) + \dfrac1{\sqrt{20}} (4,2) = \dfrac1{\sqrt{20}} (6, 6)$

Remark:$\displaystyle \dfrac1{\sqrt{20}} = \dfrac1{\sqrt{5} } \cdot \dfrac1{\sqrt{4}}$

The second angle bisector is perpendicular to the first one thus it has the direction $\displaystyle \overrightarrow{b_{\perp}} = \dfrac1{\sqrt{20}} (6, -6)$

4. You can simplify those direction vectors to $\displaystyle \vec b = (1,1)$ and $\displaystyle \overrightarrow{b_{\perp}} = (1,-1)$

5. Now use the point direction form of the equation of a line to determine the equations of the two angle bisectors.

EDIT: I've attached a sketch of the two lines and the two angle bisectors

3. Originally Posted by looi76
Question:
Find the equation of the bisector of the pair of acute angles formed by the lines $\displaystyle L_1$ and $\displaystyle L_2$ given by $\displaystyle 4x-2y-10=0$ and $\displaystyle x-2y-1=0$, respectively.
Another question! Does it make any difference if the pair formed an acute angle or obtuse angle?

4. Originally Posted by looi76
Another question! Does it make any difference if the pair formed an acute angle or obtuse angle?
No because two lines form two angles: one acute and one obtuse. With one exception: If the lines are perpendicular then the two angles are equal.

5. Originally Posted by looi76
...
Now, I need to simplify the two distance equations to get the equation of the bisector...

$\displaystyle \frac{4x-2y-15}{-2\sqrt{5}} = \frac{x-2y-1}{-\sqrt{5}}$

...
I played a little bit with your equation and much to my surprise found out that this is a very effective way to calculate the equations of the angle bisectors:

$\displaystyle \frac{4x-2y-15}{-2\sqrt{5}} = \frac{x-2y-1}{-\sqrt{5}}$ Multiply by $\displaystyle -\sqrt{20}$

$\displaystyle 4x-2y-10 = 2(x-2y-1)~\implies~y = -x+4$

$\displaystyle \frac{4x-2y-10}{-2\sqrt{5}} = \frac{x-2y-1}{+\sqrt{5}}$ Multiply by $\displaystyle -\sqrt{20}$

$\displaystyle 4x-2y-10 = -2x+4y+2~\implies~y = x-2$

Thanks for this shortcut.